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Sharygin CR P20
TheDarkPrince   37
N 4 minutes ago by E50
Source: Sharygin 2018
Let the incircle of a nonisosceles triangle $ABC$ touch $AB$, $AC$ and $BC$ at points $D$, $E$ and $F$ respectively. The corresponding excircle touches the side $BC$ at point $N$. Let $T$ be the common point of $AN$ and the incircle, closest to $N$, and $K$ be the common point of $DE$ and $FT$. Prove that $AK||BC$.
37 replies
TheDarkPrince
Apr 4, 2018
E50
4 minutes ago
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Something nice
KhuongTrang   25
N an hour ago by Zuyong
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
25 replies
KhuongTrang
Nov 1, 2023
Zuyong
an hour ago
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Equality occurs in strange points
arqady   10
N 2 hours ago by sqing
Let $a$, $b$ and $c$ are non-negatives such that $a+b+c\neq0$. Prove that:
1) \[\sum_{cyc}\sqrt{a^2+b^2}\leq\frac{11(a^2+b^2+c^2)+7(ab+ac+bc)}{3\sqrt2(a+b+c)}\]
2)\[\sum_{cyc}\sqrt{a^2+7ab+b^2}\leq\frac{2(a^2+b^2+c^2)+7(ab+ac+bc)}{a+b+c}\]
3)\[\sum_{cyc}\sqrt{4a^2+ab+4b^2}\leq\frac{5(a^2+b^2+c^2)+4(ab+ac+bc)}{a+b+c}\]
10 replies
arqady
Aug 28, 2011
sqing
2 hours ago
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An interesting inequality
JK1603JK   4
N 2 hours ago by Nguyenhuyen_AG
Source: unknown
Let a,b,c>=0 and a^2+b^2+c^2+abc=4 then prove \frac{1}{a+b+2}+\frac{1}{b+c+2}+\frac{1}{c+a+2} \le \frac{6-(a+b+c)}{4}
When does equality occur?
4 replies
JK1603JK
Mar 21, 2025
Nguyenhuyen_AG
2 hours ago
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My problem
hacbachvothuong   2
N 3 hours ago by arqady
Let $a, b, c$ be positive real numbers such that $ab+bc+ca=3$. Prove that:
$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\ge1$
2 replies
hacbachvothuong
5 hours ago
arqady
3 hours ago
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Let $a,b,c\ge 0$
khanhsy   1
N 4 hours ago by arqady
Let $a,b,c\ge 0$ and $ab+bc+ca>0$. Prove
$$\sum_{cyc}\dfrac{1}{a^3+bc(b+c)} \ge \dfrac{1}{a^3+b^3+c^3}+\dfrac{4}{ab(a+b)+bc(b+c)+ca(c+a)}.$$
1 reply
khanhsy
4 hours ago
arqady
4 hours ago
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In triangle ABC prove that:
khanhsy   0
4 hours ago
$$\dfrac{1}{cC+aB+bA}+\dfrac{1}{bB+aC+cA}+\dfrac{1}{aA+bC+cB}\ge \dfrac{1}{aA+bB+cC}+\dfrac{4}{a(B+C)+b(C+A)+c(A+B)}.$$
0 replies
khanhsy
4 hours ago
0 replies
J
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Let $a,b,c$ be nonnegative real numbers. Prove that
khanhsy   1
N 4 hours ago by KhuongTrang
Let $a,b,c,$ be nonnegative real numbers such that $a+b+c=3$. Prove that:
$$\dfrac{1}{\sqrt{a^3+3bc}}+\dfrac{1}{\sqrt{b^3+3ca}}+\dfrac{1}{\sqrt{c^3+3ab}}\ge \dfrac{1}{\sqrt{a^3+b^3+c^3+abc}}+\dfrac{2\sqrt{2}}{\sqrt{(a+b)(b+c)(c+a)}}.$$
1 reply
khanhsy
Today at 9:21 AM
KhuongTrang
4 hours ago
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An inequality on triangles sides
nAalniaOMliO   6
N Today at 7:09 AM by arqady
Source: Belarusian National Olympiad 2025
Numbers $a,b,c$ are lengths of sides of some triangle. Prove the inequality$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \geq \frac{a+b}{2c}+\frac{b+c}{2a}+\frac{c+a}{2b}$$
6 replies
nAalniaOMliO
Yesterday at 8:26 PM
arqady
Today at 7:09 AM
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Gheorghe Țițeica 2025 Grade 9 P2
AndreiVila   3
N Today at 5:06 AM by ionbursuc
Source: Gheorghe Țițeica 2025
Let $a,b,c$ be three positive real numbers with $ab+bc+ca=4$. Find the minimum value of the expression $$E(a,b,c)=\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}-(a-b)^2.$$
3 replies
AndreiVila
Yesterday at 9:09 PM
ionbursuc
Today at 5:06 AM
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Inspired by old results
sqing   7
N Today at 3:15 AM by hgomamogh
Let $ a,b,c\geq 0 $ and $a+b+c=3. $ Provethat
$$ \frac{3}{11}\leq \frac{a}{a^2+2}+\frac{b}{b^2+2}+\frac{c}{c^2+2}\leq1$$
7 replies
sqing
Yesterday at 3:10 PM
hgomamogh
Today at 3:15 AM
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J
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Geometry from germany
shinhue   3
N Jul 1, 2023 by aqwxderf
Source: German TST 2022 extension
Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP$ passes through the Euler center. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ||BC$.
3 replies
shinhue
Jun 29, 2023
aqwxderf
Jul 1, 2023
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Geometry from germany
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Reveal topic tags
geometry
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G H BBookmark kLocked kLocked NReply
Source: German TST 2022 extension
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shinhue
158 posts
shinhue
#1 h  Jun 29, 2023, 2:11 AM
Y by
Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP$ passes through the Euler center. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ||BC$.
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pi_quadrat_sechstel
582 posts
pi_quadrat_sechstel
#2 h  Jun 30, 2023, 8:03 PM • 1 Y
Y by AndresSchwepps
shinhue wrote:
Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP$ passes through the Euler center. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ||BC$.

Use complex coordinates with the excircle of $ABC$ as the unit circle. We have $o=0,h=a+b+c,p=\frac{a+b+c}{2}$. Let $R',Q'$ be the reflection of $R,Q$ in $AC,AB$ respectivly. We have
\begin{align*} r=a+b-ab\overline{p}=\frac{a+b-ab\overline{c}}{2}\\ r'=a+c-ac\overline{r}=\frac{2a+c-ac\overline{b}+c^2\overline{b}}{2}\\ y=\frac{r+r'}{2}=\frac{3a+b+c-ab\overline{c}-ac\overline{b}+c^2\overline{b}}{4} \end{align*}Analogously $z=\frac{3a+b+c-ab\overline{c}-ac\overline{b}+b^2\overline{c}}{4}$ So
\[ \frac{y-z}{c-b}=\frac{c^2\overline{b}-b^2\overline{b}}{4(c-b)}=\frac{b^2+bc+c^2}{4bc}=\frac{b\overline{c}+1+c\overline{b}}{4}\in\mathbb{R} \]So $YZ||BC$.
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AndresSchwepps
73 posts
AndresSchwepps
#3 h  Jul 1, 2023, 10:21 AM
Y by
Well, after such a short and clean complex bash, it is difficult to offer something better. I will illustrate an application of how to use the "co-tangent trick".
Let $a,b,c$ be the lenghts of the side of $ABC$, and $\alpha,\beta,\gamma$ the corresponding angles. If $X$ is the Euler center, let $\lambda=\angle QAC=\angle CAX$ and $\mu=\angle XAB=\angle BAR$. We have to prove that $\frac{AY}{AZ}=\frac{AC}{AB}$. Note that:
\begin{eqnarray*} \frac{AY}{AZ}&=&\frac{AR\cos(2\mu+\lambda)}{AQ\cos(2\lambda+\mu)}=\frac{\cos(2\alpha-\lambda)}{\cos(\alpha+\lambda)}=\\ &=&\frac{\cos(2\alpha)\cos(\lambda)+\sin(2\alpha)\sin(\lambda)}{\cos(\alpha)\cos(\lambda)-\sin(\alpha)\sin(\lambda)}=\\ &=&\frac{\cos(2\alpha)\cot(\lambda)+\sin(2\alpha)}{\cos(\alpha)\cot(\lambda)-\sin(\alpha)}. \end{eqnarray*}The problem will be solved if we prove that the above expression is equal to $\frac{b}{c}$. We need some more information about $\lambda,\mu$. Let $M,N$ be the midpoints of $AC,AB$; let $E,F$ be the feet of the altitudes of $B,C$, and let $K,N$ be the projections of $X$ onto $AC,AB$.
Let us compute $\frac{AK}{AL}$ in two different ways:
\begin{eqnarray*} \frac{AK}{AL}&=&\frac{(AM+AE)/2}{(AN+AF)/2}=\frac{b+2c\cdot \cos(\alpha)}{c+2b\cdot \cos(\alpha)},\\ \frac{AK}{AL}&=&\frac{AX\cos(\lambda)}{AX\cos(\mu)}=\frac{\cos(\lambda)}{\cos(\alpha-\lambda)}=\\ &=&\frac{\cos(\lambda)}{\cos(\alpha)\cos(\lambda)+\sin(\alpha)\sin(\lambda)}=\\ &=&\frac{\cot(\lambda)}{\cos(\alpha)\cot(\lambda)+\sin(\alpha)}. \end{eqnarray*}Equating both ratios we obtain:
\begin{eqnarray*} \cot(\lambda)(c+2b\cdot \cos(\alpha))&=&(b+2c\cdot \cos(\alpha))(\cos(\alpha)\cot(\lambda)+\sin(\alpha))\\ \cot(\lambda)(c+2b\cdot \cos(\alpha)-b\cos(\alpha)-2c\cdot \cos^2(\alpha))&=&b\sin(\alpha)+2c\cdot \cos(\alpha)\sin(\alpha)\\ \cot(\lambda)(b\cos(\alpha)-c \cos(2\alpha))&=&b\sin(\alpha)+c\sin(2\alpha)\\ b(\cot(\lambda)\cos(\alpha)-\sin(\alpha))&=&c(\cos(2\alpha)\cot(\lambda)+\sin(2\alpha)), \end{eqnarray*}as we wanted to prove.
We are still waiting for a synthetic solution.
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aqwxderf
168 posts
aqwxderf
#4 h  Jul 1, 2023, 4:36 PM • 1 Y
Y by AndresSchwepps
Let $M $ and $N $ be the midpoints of $[PR]$ and $[PQ]$.

$AMNP $- cyclic $\Rightarrow \measuredangle ANM = \measuredangle APM$.
$[AP] = [AR] \Rightarrow \measuredangle APM = \measuredangle PRA$.
$AMRY $- cyclic $\Rightarrow \measuredangle PRA = \measuredangle MYA$.
So $[MN] = [MY]$ and similarly $[MN] = [NZ]$.

Let $I $ be the orthocenter of $\triangle MAN \Rightarrow \overline{AI}$, $\overline{AP}$ - isogonal in $\measuredangle BAC$.
$\overline{MI} \perp \overline{YN}$, $[MY] = [MN] \Rightarrow \measuredangle IYN = \measuredangle INY = 90^{\circ} - \measuredangle BAC$.
Similarly $\measuredangle IZM = 90^{\circ} - \measuredangle BAC$.

Let $K $ be the reflection of $O $ over $\overline{BC}$. Its known that $K \in \overline{AP}$.
Let $L $ be the isogonal conjugate of $K $ with respect to $\triangle ABC $.
$\overline{AI}$, $\overline{AP}$ - isogonal in $\measuredangle BAC$, $K \in \overline{AP} \Rightarrow L \in \overline{AI}$.

$\measuredangle LCA = \measuredangle BCK = \measuredangle OCB = 90^{\circ} - \measuredangle BAC \Rightarrow \measuredangle LCA = \measuredangle IYN \Rightarrow IY \parallel LC$.
Similarly $IZ \parallel LB$.
$IY \parallel LC$, $IZ \parallel LB \Rightarrow YZ \parallel BC$.
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